First slide

Projection Under uniform Acceleration

Question

The speed of a projectile at its maximum height is √3/2 times its initial speed. If the range of the projectile is p times the maximum height attained by it, then p =

Moderate

A

4/3
B

2√3
C

4√3
D

3/4

Solution

$\large u\cos \theta = \frac{{\sqrt 3 }}{2}u \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{2} \Rightarrow \theta = {30^0}$
Given R = PH............ (1)
$\large \therefore Tan\theta = \frac{{4H}}{R} \Rightarrow Tan{30^0} = \frac{{4H}}{R}$
$\large \frac{1}{{\sqrt 3 }} = \frac{{4H}}{R} \Rightarrow R = 4\sqrt 3 H..........(2)$
from equation (1) & (2), we have P = 4√3

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